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Algebra anyone? 1 question -->

Algebra anyone? 1 question -->-example-1

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Hello from MrBillDoesMath!

Answer: 20 (the first choice)

Discussion:

Recall the solutions of the quadratic equation: a x^2 + bx + c = 0 are

( -b +\- sqrt ( b^2 - 4ac) ) /2a.


If there are no real solutions, then b^2 - 4ac must be < 0. In our case a = 1, b = 4, and c = c! For no real solutions this becomes

(4)^2 - 4 (1) c < 0 or

16 -4c < 0 or

16 < 4c (and dividing both sides by 4)

c > 4.


Of the choices provided only the first one, 20, is > 4.



Thank you,

MrB

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